Nodal Analysis – Dependent Voltage Source

Use nodal analysis method to solve the circuit and find the power of the $3\Omega$ – resistor.

Solution

I. Identify all nodes in the circuit.
The circuit has 3 nodes as shown below.

II. Select a reference node. Label it with the reference (ground) symbol.

The node in the middle is connected to 5 nodes and is the node with the largest number of elements connected to it. Therefore, we select it as the reference node of the circuit.

III. Assign a variable for each node whose voltage is unknown.
We label the remaining nodes as shown below. $V_1$ is connected to the reference node through a voltage source. Therefore, it is equal to the voltage of the dependent voltage source: $V_1=-6I_x$ .

IV. If there are dependent sources in the circuit, write down equations that express their values in terms of node voltages.

The voltage of the dependent voltage source is $-I_x$ . We should find this value in terms of the node voltages. $I_x$ is the current of the $5\Omega$ – resistor. The voltage across the resistor is $V_1-V_2$ . You may ask why not $V_2-V_1$ . Well, that is also correct; the voltage across the resistor is either $V_1-V_2$ or $V_2-V_1$ depend on which terminal we choose to be the positive one. In this circuit, we are going to use this voltage drop to determine $I_x$ . We prefer to use $V_1-V_2$ simply because $V_1$ is the voltage of the terminal that $I_x$ entering from. Therefore, the Ohm’s law can be applied in the simple form of $V=R \times I$ . By using the voltage drop $V_1-V_2$ , we have $I_x=\frac{V_1-V_2}{5\Omega}.$

V. Write down a KCL equation for each node.

Node of $V_1$ :
Because there is a voltage source in this node, there is no advantage in writing a KCL equation for this node. All we need to do is to use the voltage of the dependent voltage source and its relation with other node voltages:
$\left\{ \begin{array}{l} I_x=\frac{V_1-V_2}{5\Omega} \\ V_1=-6I_x \end{array} \right. \to V_1=\frac{6}{11}V_2.$

Node of $V_2$ :
$\frac{V_2-V_1}{5\Omega}+\frac{V_2}{3\Omega}-2A=0 \to -3V_1+8V_2=30.$
Substituting $V_1=\frac{6}{11}V_2$ ,
$V_2=\frac{33}{7} V \to V_1=\frac{18}{7} V.$

All node voltages are obtained. The power of the $3\Omega$ -resistor is
$R_{3\Omega}=\frac{V_2^2}{3\Omega}=7.408 W.$

The PSpice simulation result is illustrated below. The PSpice schematics can be downloaded from http://www.solved-problems.com/download/1-21.zip.

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3 Comments

Michael says:

April 22, 2012 at 4:05 pm

Thanks for the post. Very helpful.

criscolorado says:

June 15, 2012 at 3:00 am

thanx alot but i need Pspise simulation file <<and i cant download it <<

gene armstrong says:

June 17, 2013 at 1:46 am

thank you very much, it was helpful on the dependent source part of the circuit.

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Nodal Analysis – Dependent Voltage Source

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